# Find the ratio in which the line $2 x+3 y-5=0$ divides the line segment joining the points $(8,-9)$ and $(2,1)$. Also find the coordinates of the point of division.

Given:

The line $2 x+3 y-5=0$ divides the line segment joining the points $(8,-9)$ and $(2,1)$.

To do:

We have to find the ratio in which the line $2 x+3 y-5=0$ divides the line segment joining the points $(8,-9)$ and $(2,1)$ and the coordinates of the point of division.

Solution:

Let the line $2 x+3 y-5=0$ divides the line segment joining the points $A(8,-9)$ and $B(2,1)$ in the ratio $k: 1$ at point $P$.

Using section formula, we get,

$(x,y)=(\frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}ny_{1}}{m+n})$

Coordinates of $P=(\frac{k(2)+1(8)}{2+1}, \frac{k(1)+1(-9)}{2+1})$

$=(\frac{2k+8}{k+1}, \frac{k-9}{k+1})$

Point $P$ lies on the line $2 x+3y-5=0$.

This implies,

$2(\frac{2k+8}{k+1})+3(\frac{k-9}{k+1})-5=0$

$2(2k+8)+3(k-9)-5(k+1)=0$

$4k+16+3k-27-5 k-5=0$

$2k-16=0$

$k=8$

$\Rightarrow k: 1=8: 1$

Therefore, the point $P$ divides the line in the ratio $8: 1$.

The point of division $P=(\frac{2(8)+8}{8+1}, \frac{8-9}{8+1})$

$=(\frac{16+8}{9},\frac{-1}{9})$

$=(\frac{24}{9}, \frac{-1}{9})$

$=(\frac{8}{3}, \frac{-1}{9})$

Hence, the required point of division is $(\frac{8}{3}, \frac{-1}{9})$.

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